Heat Transfer Mechanisms — Question 8

Question 8

Convection and radiation act together at a heated surface

Use $\sigma=5.67\times10^{-8}\,\mathrm{W/(m^2K^4)}$. A uniform surface with area $.100\,\mathrm{m^2}$ receives steady heating power 180 W. It loses energy by convection to air at 300 K with $h=10.0\,\mathrm{W/(m^2K)}$ and by radiation to a large black enclosure at 300 K with emissivity .800. Neglect all other transfers.

1. Write the equilibrium equation and solve numerically for surface temperature.

2. At the equilibrium, find each loss contribution. Determine the local change in equilibrium temperature per extra watt of heating by differentiating the balance.

Question 8 - Solution

Convection and radiation act together at a heated surface

(a) Method and calculation

$$f(T)=hA(T-300)+\epsilon\sigma A(T^4-300^4)-180=0.$$$$f(T)=(T-300)+4.536\times10^{-9}(T^4-300^4)-180.$$$$f(400)=-.620\,\mathrm W,\quad f\prime(T)=1+1.8144\times10^{-8}T^3.$$Newton iteration $T_{n+1}=T_n-f(T_n)/f\prime(T_n)$ gives $T\approx400.28671$ K.

(b) Interpretation and further calculation

$$P_{\mathrm{conv}}=T-300\approx100.28671\,\mathrm W,\quad P_{\mathrm{rad}}=180-P_{\mathrm{conv}}\approx79.71329\,\mathrm W.$$$$\frac{dT}{dP}=\frac1{hA+4\epsilon\sigma AT^3}\approx.46217\,\mathrm{K/W}.$$The derivative is positive, so the equilibrium is unique above 300 K and a temperature increase strengthens the restoring heat loss.

Physical check

Loss mechanisms add as powers at the same surface temperature; their coefficients should not be averaged.